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6.System of Particles and Rotational Motion
hard
A uniform solid cylinder of mass $M = 3\ kg$ and radius $R = 10\ cm$ is connected about an axis through the cnetre of the cylinder to a horizontal spring with spring constant $8\ N/m$.The cylinder is pulled back, stretching the spring $1\,m$ from equilibrium.When released, the cylinder rolls without slipping. What is the speed of the center of th ecylinder when it returns to equilibrium? .................. $m/s$
A$1.33$
B$1$
C$1.15$
D$1.41$
Solution
$\mathrm{w}_{\mathrm{sp}}=\Delta \mathrm{k}$
$\frac{1}{2} \mathrm{kx}^{2}=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \times \frac{1}{2} \mathrm{mR}^{2} \times \frac{\mathrm{v}^{2}}{\mathrm{R}^{2}}=\frac{3}{4} \mathrm{mv}^{2}$
$v=\sqrt{\frac{2}{3} \frac{k}{m} X^{2}}=\sqrt{\frac{2 \times 8}{3 \times 3}} \times 1=\frac{4}{3}$
$\frac{1}{2} \mathrm{kx}^{2}=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \times \frac{1}{2} \mathrm{mR}^{2} \times \frac{\mathrm{v}^{2}}{\mathrm{R}^{2}}=\frac{3}{4} \mathrm{mv}^{2}$
$v=\sqrt{\frac{2}{3} \frac{k}{m} X^{2}}=\sqrt{\frac{2 \times 8}{3 \times 3}} \times 1=\frac{4}{3}$
Standard 11
Physics
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